If function f(x) given by
`f(x)={{:(,(sin x)^(1//(pi-2x)),xne pi//2),(,lambda,x=pi//2):}` is continous at `x=(pi)/(2)` then `lambda`=

Options

A. e
B. 1
C. 0
D. none of these

Explanation

To determine the value of \( \lambda \) such that the function

\[ f(x) = \begin{cases} (\sin x)^{\frac{1}{\pi - 2x}} & \text{if } x \neq \frac{\pi}{2} \\ \lambda & \text{if } x = \frac{\pi}{2} \end{cases} \]

is continuous at \( x = \frac{\pi}{2} \), we need to ensure that the limit of \( f(x) \) as \( x \) approaches \( \frac{\pi}{2} \) is equal to \( f\left(\frac{\pi}{2}\right) \).

### Step 1: Find \( \lim_{x \to \frac{\pi}{2}} f(x) \)

We first need to calculate the limit:

\[ \lim_{x \to \frac{\pi}{2}} (\sin x)^{\frac{1}{\pi - 2x}} \]

As \( x \) approaches \( \frac{\pi}{2} \), \( \sin x \) approaches \( 1 \). Therefore, we can rewrite the limit as:

\[ \lim_{x \to \frac{\pi}{2}} (\sin x)^{\frac{1}{\pi - 2x}} = 1^{\text{something}} = 1 \]

However, we need to analyze the exponent \( \frac{1}{\pi - 2x} \) as \( x \) approaches \( \frac{\pi}{2} \).

### Step 2: Analyze the exponent

As \( x \to \frac{\pi}{2} \), \( \pi - 2x \) approaches \( 0 \). Thus, we need to evaluate the behavior of \( \frac{1}{\pi - 2x} \):

\[ \lim_{x \to \frac{\pi}{2}} \frac{1}{\pi - 2x} = \lim_{x \to \frac{\pi}{2}} \frac{1}{\pi - 2\left(\frac{\pi}{2}\right)} = \lim_{x \to \frac{\pi}{2}} \frac{1}{0} \]

This limit approaches \( \infty \) as \( x \) approaches \( \frac{\pi}{2} \) from the left and approaches \( -\infty \) from the right.

### Step 3: Evaluate the limit

Now we can evaluate the limit:

\[ \lim_{x \to \frac{\pi}{2}^-} (\sin x)^{\frac{1}{\pi - 2x}} \to 1^{+\infty} = 1 \]

\[ \lim_{x \to \frac{\pi}{2}^+} (\sin x)^{\frac{1}{\pi - 2x}} \to 1^{-\infty} = 0 \]

### Step 4: Set the limits equal to \( \lambda \)

For the function to be continuous at \( x = \frac{\pi}{2} \):

\[ \lambda = \lim_{x \to \frac{\pi}{2}} f(x) = 1 \]

Thus, we find that \( \lambda = 1 \).

### Conclusion

The value of \( \lambda \) such that the function \( f(x) \) is continuous at \( x = \frac{\pi}{2} \) is

\[ \lambda = 1 \]